python之查找列表中的重复元素

我有一个列表:

nums = [1, 2, 3, 1, 5, 2, 7, 11] 

我正在尝试创建一个函数，返回每个数字在列表中出现的次数。输出可能如下所示:

1 occurred 2 times 
2 occurred 2 times 
3 occurred 1 time 
5 occurred 1 time 
... 
... 

这是我到目前为止尝试过的:
-为列表中的每个元素创建字典
-让嵌套循环遍历每个元素并对照每个其他元素检查它
-如果元素匹配，则向该元素的字典键加一

问题:
每次循环时，它都会重新检测相同的元素。因此无论某元素有多少，不是加1，而是提高n次方

Enter integers between 1 and 100: 5 2 41 4 5 2 2 4 
4 occurs 4 times 
2 occurs 9 times 
41 occurs 1 times 
5 occurs 4 times 

代码:

def main(): 
    original_nums = input("Enter integers between 1 and 100: ") 
    nums = [i for i in original_nums.split()] 
    my_dict = {} 
 
    for i in nums: 
        my_dict[i] = 0 
 
    for i in nums: 
        for j in nums: 
            if i == j: 
                my_dict[i] += 1 
 
    for i in my_dict: 
        print(i,"occurs",my_dict[i],"times") 
 
 
 
if __name__ == "__main__": 
    main() 

请您参考如下方法:

Counter就是你需要的一切

>>> from collections import Counter 
>>> Counter([1, 2, 3, 1, 5, 2, 7, 11]) 
Counter({1: 2, 2: 2, 3: 1, 5: 1, 7: 1, 11: 1}) 

或者您可以像这样修复您的代码

def main(): 
    original_nums = input("Enter integers between 1 and 100: ") 
    nums = [i for i in original_nums.split()] 
    my_dict = {} 
 
    for i in nums: 
        my_dict[i] = my_dict.get(i, 0) + 1 
        # or .setdefault(i, 0) instead of .get(i, 0) 
        # setdefault is generally faster 
 
    for i in my_dict: 
        print(i, 'occurs', my_dict[i], 'times') 
 
if __name__ == '__main__': 
    main() 

运行时:

Enter integers between 1 and 100: 5 5 5 5 1 2 3 3 3  
1 occurs 1 times 
2 occurs 1 times 
3 occurs 3 times 
5 occurs 4 times